Q1. The sum of ages of family members (both children and parents) is 360 years.The total ages of children and parents are in the ratio 2:1 and the ages of wife and husband are in the ratio 5:7.What will be the age of husband?

A. 65 B. 75 C. 72 D. 70

Answer: Option D
Explanation: Sum of ages is 360 years.The ratio of children and parents ages is 2:1. Total age of parents = 360 x [latex]\frac{1}{3}[/latex] = 120 years Ratio of wife and husband age is 5:7. Therefore, the age of husband = 120 x [latex]\frac{7}{12}[/latex] = 70 Hence the age of husband is 70 years.
2. The ratio between the present ages of A and B is 6:7. If B is 4 years old than A. What will be the ratio of the ages of A and B after 4 years?

A. 5 : 7 B. 7 : 8 C. 8 : 7 D. Data inadequate

Answer: Option B
Explanation: Let A’s age and B’s age be 6X years and 7X years respectively Then, 7X – 6X = 4 X = 4 Required ratio = (6X + 4) : (7X + 4) = (6*4 + 4) : (7*4 + 4) =28 : 32 = 7 : 8
3. Sush was thrice as old as Poonam 6 years back. Sush will be 5/3 times as old as Poonam 6 years hence. How old is Poonam today?

A. 20 B. 24 C. 28 D. 35

Answer: Option C
Explanation: Let Poonam age 6 years back = x. Then, Sush age 6 years back = 3x. (5/3) * (X + 6 + 6) = (3X + 6 + 6) So 5(x + 12) = 3(3x + 12), so x = 6. Poonam Age = (x + 6) years = 12 years
4. Present ages of Simmi and Anu are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively, What is Anu’s present age in years?

A. 20 B. 40 C. 15 D. 20

Answer: Option B
Explanation: Let the present ages of Simmi and Anu be 5x years and 4x years respectively Then, [latex]\frac{[5X + 3]}{[4X + 3]}[/latex] = [latex]\frac{11}{9}[/latex] x = 6 Anu’s present age = 4X = 24 years
5. If 6 years are subtracted from the present age of Anuj and the remainder is divided by 18, then the present age of his grandson Gopal is obtained.If Gopal is 2 years younger to Mohan whose age is 5 years, then what is the age of Anuj?

A. 44 B. 60 C. 80 D. 92

Answer: Option B
Explanation: Let Anuj’s age be X Gopal is 2 years younger than Mohan, so Gopal is 3 years (i.e 5 – 2 = 3) If Arun had born 6 years before, his age would had been X – 6. As per the question, X – 6 should be 18 times as that of Gokul’s age. i.e. [latex]\frac{(X – 6)}{18}[/latex] = 3 X - 6 = 3 x 18 x = 60

Q1. 3 years ago, the average of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is:

A. 5 years B. 2 years C. 1 year D. 4 years

Answer: Option B
Explanation: Total age of 5 members, 3 years ago = (17 x 5) years = 85 years Total age of 5 members now = (85 + 3 x 5) years = 100 years Total age of 6 members now = (17 x 6) years = 102 years Age of the baby = (102 – 100) years = 2 years
2. Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is:

A. 45 B. 25 C. 30 D. 45

Answer: Option D
Explanation: Let the fourth number be a then hen, third number = 5a, second number = [latex]\frac{5a}{3}[/latex] and first = [latex]\frac{10a}{3}[/latex] a+ 5a+ [latex]\frac{5a}{3}[/latex] + [latex]\frac{10}{3}[/latex] = 24.75 x 4 so, a = 9 So, the numbers are 9, 45, 15 and 30
3. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is:

A. 1 : 4 B. 2 : 3 C. 3 : 4 D. 4 : 2

Answer: Option B
Explanation: Let the ratio be k:1 Then, k x 16.4 + 1 x 15.4 = (k + 1) x 15.8 (16.4 – 15.8) k = (15.8 – 15.4) k = [latex]\frac{2}{3}[/latex]
4. The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books?

A. 50 B. 44 C. 45 D. 42

Answer: Option C
Explanation: Total ages of 30 boys = 14 x 30 = 420 years Total age when class teacher is included = 15 x 31 = 465 years Age of class teacher = 465 – 420 = 45 years
5. The average age of 30 boys of a class is equal to 14 years. When the age of the class teacher is included the average becomes 15 years. Find the age of the class teacher?

A. 44 B. 60 C. 80 D. 92

Answer: Option B
Explanation: Let Anuj’s age be X Gopal is 2 years younger than Mohan, so Gopal is 3 years (i.e 5 – 2 = 3) If Arun had born 6 years before, his age would had been X – 6. As per the question, X – 6 should be 18 times as that of Gokul’s age. i.e. [latex]\frac{(X – 6)}{18}[/latex] = 3 X - 6 = 3 x 18 x = 60

1. The area of a rectangular field is 3375 [latex]{m}^{2}[/latex] and its length and breadth are in the ratio 5:3. Find the breadth ?

A. 125m B. 45m C. 225m D. 65m

Answer: Option B
Explanation: a + b = 6 5x × 3x = 3375 15[latex]{x}^{2}[/latex] = 3375 [latex]{x}^{2}[/latex] = [latex]\frac{3375}{15}[/latex] = 225 x = 15 3x = 3*15 = 45
2. The perimeter of a square is twice the perimeter of a rectangle. If the perimeter of a square is 48cms and the length of the rectangle is 7cm. Find the breadth of the rectangle?

A. 5cm B. 4cm C. 9cm D. 5

Answer: Option A
Explanation: P of Square = 4a = 48 A = [latex]\frac{48}{4}[/latex] = 12cm P of rectangle = [latex]\frac{48}{2}[/latex] = 24cm = 2( l + b) 2(7 + b) = 24 B = 12-7 = 5
3. The area of a square is 7sqm less than half of the area of a rectangle. The length of the rectangle is 14m and breadth is 6m less than its length. What is the perimeter of the square ?

A. 42cm B. 32cm C. 22cm D. 28cm

Answer: Option A L = 14, b = 8 L*b = 14*8 = 112 Area of square = ([latex]\frac{112}{2}[/latex]) – 7 = 56-7 = 49 A = 7 Perimeter = 4a = 4*7 = 28cm
4. A cow is tethered to a peg with 7m long rope at the corner of a 35m long and 20m wide rectangular grass field, What area of the field will the cow graze ?

A. 38.47 [latex]{m}^{2}[/latex] B. 35.64 [latex]{m}^{2}[/latex] C. 43.65 [latex]{m}^{2}[/latex] D. 40[latex]{m}^{2}[/latex]

Answer: Option A
Explanation: A = [latex]\frac{1}{4}[/latex](3.14r2) = [latex]\frac{3.14*7*7}{4}[/latex] = 38.47[latex]{m}^{2}[/latex]
5. Inside a square plot a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is 14m. What is the area of space left out in the square plot after developing the garden ?

A. 32 [latex]{m}^{2}[/latex] B. 40 [latex]{m}^{2}[/latex] C. 42 [latex]{m}^{2}[/latex] D. 35[latex]{m}^{2}[/latex]

Answer: Option C
Explanation: area of space left = 14*14 – (3.14*7*7) = 196 – 153.86 = 42.14 = 42[latex]{m}^{2}[/latex]